fault current calculations - university of nevada, las vegaseebag/fault current calculations.pdf ·...

Fault Current Calculations

ECG 740

Introduction

• A fault in a circuit is any failure that interferes with the normal system operation.

• Lighting strokes cause most faults on high-voltage transmission lines producing a very high transient that greatly exceeds the rated voltage of the line.

• This high voltage usually causes flashover between the phases and/or the ground creating an arc.

• Since the impedance of this new path is usually low, an excessive current may flow.

• Faults involving ionized current paths are also called transient faults. They usually clear if power is removed from the line for a short time and then restored.

Introduction

• If one, or two, or all three phases break or

if insulators break due to fatigue or

inclement weather, this fault is called a

permanent fault.

• Approximately 75% of all faults in power

systems are transient in nature.

• Knowing the magnitude of the fault

current is important when selecting

protection equipment (type, size, etc..)

3-Phase fault current transients in synchronous generators

When a symmetrical 3-phase fault

occurs at the terminals of a

synchronous generator, the resulting

current flow in the phases of the

generator appear as shown.

The current can be represented as a

transient DC component added on top

of a symmetrical AC component.

Before the fault, only AC voltages and

currents are present, but immediately

after the fault, both AC and DC

currents are present.

Symmetrical AC component of the fault current:

• There are three periods of time: – Sub-transient period: first couple of cycles after the fault – AC

current is very large and falls rapidly; – Transient period: current falls at a slower rate; – Steady-state period: current reaches its steady value.

• It is possible to determine the time constants for the sub-transient and transient periods .

Fault current transients in machines

• The AC current flowing in the generator during the sub-transient period is called the sub-transient current and is denoted by I”. The time constant of the sub-transient current is denoted by T” and it can be determined from the slope. This current can be as much as 10 times the steady-state fault current.

• The AC current flowing in the generator during the transient period is called the transient current and is denoted by I’. The time constant of the transient current is denoted by T’. This current is often as much as 5 times the steady-state fault current.

• After the transient period, the fault current reaches a steady-state condition Iss. This current is obtained by dividing the induced voltage by the synchronous reactance:

Fault current transients in machines

• The rms value of the AC fault current in a synchronous generator varies over time as

• The sub-transient and transient reactances are defined as the ratio of the internal generated voltage to the sub-transient and transient current components:

""

AEX

I

" '" ' 't T t T

ss ssI t I I e I I e I

''

AEX

I A

ss

s

EI

X

Fault current calculations

Example 1: A 100 MVA, 13.8 kV, Y-connected, 3 phase 60 Hz synchronous

generator is operating at the rated voltage and no load when a 3 phase fault occurs

at its terminals. Its reactances per unit to the machine’s own base are

1.00 ' 0.25 " 0.12sX X X

and the time constants are

' 1.10 " 0.04T s T s

The initial DC component averages 50% of the initial AC component.

a) What is the AC component of current in this generator the instant after the fault? I” = 34,900 A.

b) What is the total current (AC + DC) in the generator right after the fault occurs?

c) What will the AC component of the current be after 2 cycles? After 5 s?

1.5 " 52,350totI I A

17,910 12,142 4,184 24,236

30I A

5 0 133 4,184 4,317I A

Symmetrical fault current calculations

• To determine the fault current in a large power system: – Create a per-phase per-unit equivalent circuit of the

power system using either sub-transient reactances (if subtransient currents are needed) or transient reactances (if transient currents are needed).

– Find the Thevenin equivalent circuit looking from the fault point, then divide the Thevenin voltage by the Thevenin impedance.

• You may use the Z-bus elements to determine the voltages and current flows elsewhere in the system (further away from the faulted bus)

• Today, software tools to do the calculations for us are readily available (e.g. Powerworld).

Fault current calculations using the impedance matrix

Before the fault, the voltage on bus 2 was Vf. If

we introduce a voltage source of value Vf

between bus 2 and the neutral, nothing will

change in the system.

Since the system operates normally before the

fault, there will be no current If” through that

source.


Assume that we create a short circuit on bus 2,

which forces the voltage on bus 2 to 0. This is

equivalent to inserting an additional voltage

source of value -Vf in series with the existing

voltage source. The later will make the total

voltage at bus 2 become 0.

With this additional voltage source, there will be

a fault current If”, which is entirely due to the

insertion of the new voltage source to the

system. Therefore, we can use superposition to

analyze the effects of the new voltage source on

the system.

The resulting current If” will be the current for the

entire power system, since the other sources in

the system produced a net zero current.


If all voltage sources except –Vf” are set to

zero and the impedances are converted to

admittances, the power system appears as

shown.

For this system, we can construct the bus

admittance matrix as discussed previously:

16.212 5.0 0 6.667

5.0 12.5 5.0 2.5

0 5.0 13.333 5.0

6.667 2.5 5.0 14.167

bus

j j j

j j j jY

j j j

j j j j

The nodal equation describing this power system is

busY V = I


With all other voltage sources set to zero, the voltage at bus 2 is –Vf, and the

current entering the bus 2 is –If”. Therefore, the nodal equation becomes

11 12 13 14 1

"

21 22 23 24

31 32 33 34 3

41 42 43 44 4

0

0

0

f f

Y Y Y Y V

Y Y Y Y V I

Y Y Y Y V

Y Y Y Y V

where V1, V3, and V4 are the changes in the voltages at those busses due to the

current –If” injected at bus 2 by the fault.

The solution is found as -1

bus busV = Y I = Z I


11 12 13 141

"

21 22 23 24

31 32 33 343

41 42 43 444

0

0

0

f f

Z Z Z ZV

Z Z Z ZV I

Z Z Z ZV

Z Z Z ZV

Which, in the case considered, is

where Zbus = Ybus-1. Since only bus 2 has current injected at it, the system reduces to

"

"

1 12

"

3 32

"

2

4

2

42

f

f

f

f f

V Z I

V Z I

V

I

Z I

V Z


Therefore, the fault current at bus 2 is just the prefault voltage Vf at bus 2 divided by

Z22, the driving point impedance at bus 2.

"

22

f

f

VI

Z

The voltage differences at each of the nodes due to the fault current can be

calculated by substitution:

" 121 12

22

2

" 323 32

22

" 424 42

22

f f

f f

f f

f f

ZV Z I V

Z

V V V

ZV Z I V

Z

ZV Z I V

Z

Symmetrical Components and Unsymmetrical Faults

Example 1:

Last Assignment

• Watch the 20 min video on how to navigate Powerworld when conducting fault analysis at the link below, then calculate the 3-phase and single-phase fault currents at each of the 6 busses and resulting voltages.

https://www.powerworld.com/training/online-training/fault-analysis







fault current calculations - university of nevada, las vegaseebag/fault current calculations.pdf ·...

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